For the decomposition reaction
`NH_2COONH_4 (s) hArr 2NH_3(g) +CO_2(g)`
The `K_p=2.9 xx 10^(-5) atm^3`. The total pressure of gases at equilibrium when 1 mole of `NH_2COONH_4` (s) was taken to start with would be

Options

A. 0.0194 atm
B. 0.0388 atm
C. 0.0582 atm
D. 0.0766 atm

Answer

C

Explanation

`{:(NH_2COOHN_4(s), hArr 2NH_3(g), +CO_2(g)),(1,2,1):}`
`K_p=2.9xx10^(-5)atm^3`
If P is the total pressure at equilibrium
`K_p=((2p)/3)^2(p/3)`
`therefore P_3=(27xx2.9xx10^(10^(-5)))/4=1.9575`
`P=3sqrt(1.9575)=0.0582`